wanminliu@gmail.com
§
16.1
Vector
and
scalar
fields
5=1
.
Sketch
.
H
'
IR
}
Vertov
field
F-
10.07=10.0
)
F-
:
IR
}
→
IR
}
F
(
1.
07=(0-1)
(
✗
'
%
Z
)
(
F
,
E
,
F
,
)
F
(0-1)
=
(
1,0
)
fieldfares
.
plane
vector
field
F
:
CR
'
→
IR
'
4¥
=
¥÷
(
X
,y
)
(
→
(
Fi
,
Fz
)
Fi
=
Fi
1×14
)
1=1 .2
.
⇐
÷
"
"
"
"
"
¥=¥
#
3
Sketch
the
given
plane
vector
field
and
determine
its
field
lines
.
fxdx-fydyT-cx.gg
=
yI+xJ
=
(
F
'
'
E)
Ix
'
=
z
'
y'
+
§
.
=
(
y
,
X
)
EIR
'
C.
=/
X
?
>
-
=/
C
=
-
I
✗
"
-
Y
'
=
-1
wanminliu@gmail.com
F
,
=
e×
E=
e-
✗
#
5
Same
question
as
#
3
.
Fix
.y
,
=
exit
e-
✗
J
¥
=
d¥
Field
lines
equation
e-
"
dx
=
dy
¥±=¥i
=
d÷
3-
dim
(E)
I
e-
"
dead
=/
dy
y
=
-
I
e-
"
+
c.
9¥
=
d¥
2-
dim
☐
Sol
.
Fix
,y
,
=
(
et
,
e'
×
)
£ieue
"
wanminliu@gmail.com
https://en.wikipedia.org/wiki/Archimedean_spiral
#
17
Determine
the
field
lines
of
the
Sol
.
Frfr
.
0
)
=
1
given
polar
vector
fields
.
Fo
Cr
,
O
)
=
r
É=É+rÉ
÷
=
rd÷
=
do
Recall
É=
coso
It
Sino
J
É
=
-
Sinai
-1
cos
of
r
=
0
+
c.
4
court
.
É
=
EIKO
)=
Frfr
.
01%-1
Foer
.
0
)É
Remark
:
r
=
atbo
for
a.
b
EIR
constants
Field
lines
equation
in
the
polar
coordinate
,
(61--0)
rdo
then
its
trace
is
called
Archimedean
spiral
¥÷
=
To
⇐
"
"
"
"
r=
-0
wanminliu@gmail.com
§
16.2
Conservative
fields
"
a
vector
field
that
is
coming
from
a
scalar
field
Necessary
condition
more
precisely
,
the
gradient
of
the
scalar
field
.
"
Def
-
If
F-
IX.
7.
2)
=
00/1×14.2-1
.
(
for
a
conservative
plane
vector
field
)
=
(
0¥
,
0¥
,
3¥
)
É
ix.
7)
=
F.
ix.
4)
it
F.
ix.
y
,
J
§i
then
the
condition
U
=
;¥ix¥J+°¥ñ
2¥;
=
÷
Fil
"
'D
in
a
domain
D.
then
the
vector
field
É
•
(
3d
version
)
is
caned
a
conservative
vertor
f-
held
in
D
,
F-
IX.
7.
2)
=
F
,
ix.
7.
2)
Tt
Fak
.
y.tn?--pglx.y.z)E
and
the
fountain
01
is
called
a
(
scalar
)
poten tial
then
¥
,=¥
.
for
f-
on
D
.
0/-1
Court
.
is
also
a
potential
4-
=
1¥
814+4=04
.
3¥
=
IF
wanminliu@gmail.com
#
3
Determine
whether
the
giv en
field
is
conservative
,
and
find
a
potential
if
it
is
.
(
dy
±
d
É
is
not
conservative
.
É
☒
=
Hint
.
check
±
.
÷
.
with
F
,
IX.
Y
)
=
¥y
,
F
,
LX
-
Y
)
=
-
¥72
.
☐
wanminliu@gmail.com
#
7
Find
the
three
dim
.
vector
tied
so
tx-3-xcx.x.li#y----i
)
with
potential
✗
'
=
1¥
,
,
=
"
'
fg.x.ie#y.+*+,y-z2H-H-2(X-Xo
)
Sol
.
=
(
✗
.
y
.
z
)
C-
IR
}
=
É
É
=
(
Xo
.
Yo
,
to
)
c-
IR
}
,
a
fixed
similarly
vector
.
¢
,
=
;¥¥?÷
IF
-
To
/
'
=
CX -xoi-ly-yoj-lt-t.is
*
=
We
wan t
to
compute
THE
)=(¢×
,
4
,
,
¥1
841M
=
(
x-xo.Y-Y.it
-
to
)
41×1%2-1
=
☒.×¥É+¥
=
→(rÑ
)
H
-
r
:p
☐
wanminliu@gmail.com
#
21
show
that
①
✗
=
{
r
'
sinzo-c.IO
)
③
É=
r
5m20
É
+
r
cos
20
É
is
conservative
,
and
find
a
potential
for
it
.
②
2¢
To
=
r
'
cos
20
Sol
.
We
wan t
to
find
a
function
Plug
③
Into
.
we
have
✗
=
cfcr
,
G)
So
that
trccoszo
)
.
2
+
c.
'
(
O
)
=
Ñ
who
I
§
(
r
,
O
)
=
¥
=
I
F.
1h01
.
Filho
'
)
c
,
'
(
O
)
=
r
'
cos
20
-
Hanzo
=
0
=
(
r
sin
20
,
run
20
)
C
,
(
O
)
=
C
.
Kennard
:
Polino
)=¥r
Et
f-
I¥É
§
=
Ir
'
since
-1
C
so
we
want
to
solve
0¥
=
rsinzo
①
It
is
a
potential
of
=p
.
,
÷¥o
=
rozo
①
wanminliu@gmail.com
§
16.3
line
integrals
Ñ=(
etat
,
etsint
,
t
)
Sol
.
d÷=(
etast-etts.int/,etsTnt+etwit
,
1)
#
7
Find
the
moment
of
inertia
about
the
Z-axis
,
i.
e.
the
value
of
t.gr/=etwst-etsint)Iets-mt-etcort)--#L=crfetx7-y4ds
=
eeÉ
=
ze'#
for
a
wire
of
constant
density
dbiYz-orfYf@tu.t
,
I
Cetsintj
)2é#
dt
along
the
wine
e
:
=D
/
o
"
eat
zé
at
F
=
etwsti-etsintj-tk-t.ES?-zeit-Td(zeit-i
)
from
t
=
0
to
-1=27
.
$=zet
C-
frs
,
25€
]
2ei
FACT
:
{
fix
-7
.
2-
ids
=
fab
firm
,
/
¥1
/
lot
=
If
Jr
,
s
ds
'
=
¥
.2f↳¥Tdg
}
Eds
B
3-
after
=
§
e.
%)
?
-
}
?
]
.
pg
wanminliu@gmail.com
Ñ
=
(
cost
,
Sint
,
t
)
#
11
Find
the
mass
and
centre
of
mass
drop
a
wire
bent
in
the
shape
of
the
at
=
f-
Sint
,
wit
.
1)
circular
helix
µ÷
/
=
ÉÉ
=
F
✗
=
Cost
,
Y
=
Sint
,
Z
=
t
,
often
Mass
if
the
wire
has
line
density
given
by
m
=
f
fix
-
ya
,
Is
=
f
2-
Is
e
e
J
IX.
Y
,
Z
)
=
2-
.
21T
=/
o
t.rdt-f-t.to
"
Sol
.
FACT_
:
mass
=
f
fix
.az
>
ds
e
=
zrzt i
.
Center
of
mass
f
centroid
)
f¥"""!e;÷;÷ !¥
wanminliu@gmail.com
We
compute
centroid
.
integration
by
parts
§
✗
fix
.
yards
=/
"
'
cost
.
trzdt
=
f
f
?
"
cost
)
+
at
☐
=
R
lo
"
t
dfsint
)
=
if
¢
.
sint
/
"
-
f.
"
csint
)
It
]
0
=
Jz
[
27
.
5in
27
-
O
-
5in
0
-
f%5#dt
]
=
0
.
0
fey
dixie
.
2-
ids
=
f
"
sint
.
t
it
dt
=
1-
f)
fo
"
t
dirt
)
=
try
[
t
-
cost
/
?
-
Hutt
]
=
try
/
a.
1-
0
.
D=
-
Vix
{
2-
dixie
.
Hds
=
f)
"
c-
'
fat
=
§
Izzy
}
Now
the
centroid
=
l°'-¥¥¥"-
=
10
,
-
¥
,
4¥
)
☒
wanminliu@gmail.com
§
16.4
line
integrals
of
Th
1
(
Independence
of
path
)
vector
fields
Let
D
be
an
open
,
connected
domain
,
and
let
É
be
a
smooth
vector
field
on
D
.
I
É
.
di
Then
the
following
three
statements
are
E
equivalent
É
=
(
E
,
E
,
F
}
)
(a)
É
is
conservative
on
D.
i.
e
-7=08
dÑ
=
(
dx
,
dy
,
d
2-
)
(b)
§
É
.
dñ
=
o
ftp.ieeulu
smooth
e
Ee
on
D
(
starting
point
=
ending
point
)
For
smooth
are
(c)
Given
any
two
points
R
&
P
,
ink
Nothing
to
do
With
the
notion
of
}
=
flt
)
=
(
✗
It
,
y
16-1
,
2-
It
)
)
{
É
di
"
closed
"
in
top
.by
a
ft
a-
b
has
the
same
value
for
any
e
piecewise
smooth
curve
slating
at
Po
and
{
É
.
dñ
=
f
?
/
É
.
d¥)dt
ending
at
R
.
=
Jab
[
É
1×1+1
.
Yeti
,
titi
)¥+
+
É
1×1+ 1
.
>
it
,
2-
it
)
¥
+
ÉIXHI
.
Yet
!
2-
it
)¥+
]
dt
wanminliu@gmail.com
Evaluate
the
line
integral
of
the
tang ent ial
component
of
the
given
vector
field
along
the
given
curves
.
#
I
ɵ
,
y
)=(
xy
,
-
×
'
)
E
:
Y=x
'
from
10.0 7
to
(
1.
1)
f
É
.
dt
=
f
ki
,
E)
fdx
,
dy
)
e
e
=
{
F.
dx
+
Tidy
=
f
!
kid
✗
+
f.
'
C-
✗
9
=
-
lo
'
x3dx=
-
§
ZXIX
Now
e
can
be
para
retired
by
✗
{
X=x×e[÷
this
is
the
☐
"
+
"
variable
y
=
✗
2
wanminliu@gmail.com
#
5
Fix
,y
.
zj
=
(
Yt
,
xt
,
✗
Y
)
How
to
parametric
such
curve
?
from
f-
1.
o.o
)
to
11.0.0
)
✗
2+5=1
{
2-
=
y
along
either
dlration
of
the
curve
of
✗
=
cos
0
Since
we
want
✗
changes
intersection
of
the
cylinder
✗
4-
y
'
=L
from
-1
to
r
y
=
5in
0
and
the
plane
2-
=L
(
because
1-1
,
0.9
to
(
leg
)
so
we
take
the
rage
of
0
C-
[
a.
22
]
blue
#
we
:*
we
curve
"
"
"
with
the
help
of
Flo
!
we
can
do
-06
since
4=51
no
computation
,
runs
from
we
can
also
chose
0
to
-1
,
to
0
qÑl01=(5n0,wo,
I
9
Off
-
E.
E
]
a
3¥
27
wanminliu@gmail.com
You
can
also
try
he
came
/
blue
curve
÷%y
pµ,=1gzo,-↳o,.a
✗
Q
line
segment
Of
[
-
E
,
I
]
let
us
take
the
cane
POT
É=yowtao=
6
Then
{
¥
.
di
=
f
I.
dr
=
/
I
dr°=
f
o
.
dñ
Any
smooth
⑧
Ñ
cure
from
P
to
a
/
did
=\
DX
,
0,0
)
so
Vertov
field
=P
is
conservative
.
Yeo
=
t.io
.d×=
0
.
2-
to
✗
c-
FI
.
11
☐
wanminliu@gmail.com
so
we
let
#
T
Find
the
work
done
by
the
10=-1×4×-1
-
Yz
+
It
'
fore
field
E
,
E
g
it
then
satisfies
①
②
③
.
-7=1×+4
,
X
-
Z
,
2--7
)
so
É
is
a
conservative
field
.
In
moving
an
object
from
11
,
o
,
-11
to
/
0
,
-
2.
3)
aknganysmoohcur#
Then
¥
We
want
to
know
if
É
is
conservative
.
fe-I.DE
=
0/10
.
-2, 3
)
-
✗
(
c.
°
.
-1
)
(
You
can
check
the
compatibility
conditions
:
3=4--37×-1
¥z=¥×
.
¥z=¥y
)
.
=
-
C-
4×3+-21×9
-
-1×12
-
Éxlti
Suppose
É=
to
for
some
10=101×1%2-1
,
then
=
6+-9
-
É
-
É
=
9.5
.
Oh
,
=
✗
+
y
✗
=
f-
✗
'
+
xy
-1
quiz
)
①
toy
=
x
-
2-
4=1×-2-17-1
914 2-1
③
^
☐
✗
z
=
2-
-
y
of
=
d-
F-
zy
+
g)
ix.
7)
①
wanminliu@gmail.com
#
15
Evaluate
(b)
f.
"
asinodlawo
)
eh
§
✗
dy
6)
§
Ydx
e
=
-
a2µsin
e
t
along
the
closed
curve
e
=
circle
✗
'
+
{
=
a
2
.
1¥
osI
do
=
I
-
✗
aero
•
→
az
y=asino
0
C-
(
°
'
"
]
☐
→
✗
27
f.
aero
dlasino
)
=
a
'
f.
"
wie
do
Remark
.
The
role
of
✗
and
y
=
a
'
f.
"
Ilario
-11
)
do
ar e
NOT
symmetric
with
respect
to
=
a
'
.
-21
(
2k
+
I
/
F'
mzodlo
)
the
circle
since
the
circle
has
a
direction
.
By
default
,
we
call
the
counterclockwise
direction
as
positiv e
=
-×
a
'
direction
.
